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BJT Ampliﬁer Circuits

As we have developed diﬀerent models for DC signals (simple large-signal model) and AC signals (small-signal model), analysis of BJT circuits follows these steps: DC biasing analysis: Assume all capacitors are open circuit. Analyze the transistor circuit using the simple large signal mode as described in pages 77-78. AC analysis: 1) Kill all DC sources 2) Assume coupling capacitors are short circuit. The eﬀect of these capacitors is to set a lower cut-oﬀ frequency for the circuit. This is analyzed in the last step. 3) Inspect the circuit. If you identify the circuit as a prototype circuit, you can directly use the formulas for that circuit. Otherwise go to step 4. 4) Replace the BJT with its small signal model. 5) Solve for voltage and current transfer functions and input and output impedances (nodevoltage method is the best). 6) Compute the cut-oﬀ frequency of the ampliﬁer circuit. Several standard BJT ampliﬁer conﬁgurations are discussed below and are analyzed. For completeness, circuits include standard bias resistors R1 and R2 . For bias conﬁgurations that do not utilize these resistors (e.g., current mirror), simply set RB = R1 R2 → ∞. Common Collector Ampliﬁer (Emitter Follower) DC analysis: With the capacitors open circuit, this circuit is the same as our good biasing circuit of page 110 with RC = 0. The bias point currents and voltages can be found using procedure of pages 110-112. AC analysis: To start the analysis, we kill all DC sources:
VCC = 0 R1 vi Cc vo R2 RE R1 R2 vi Cc B E RE C vo

VCC R1 vi Cc vo R2 RE

ECE65 Lecture Notes (F. Najmabadi), Winter 2006

123

We can combine R1 and R2 into RB (same resistance that we encountered in the biasing analysis) and replace the BJT with its small signal model:
vi Cc

B
+ ∆v
BE

∆i

B

∆i β ∆i
B

C

C

vi

Cc

B
∆i
B

rπ

E
RE β ∆i
B

vo

rπ

ro vo

RB

RB

_

E

ro

RE

C

The ﬁgure above shows why this is a common collector conﬁguration: collector is shared between input and output AC signals. We can now proceed with the analysis. Node voltage method is usually the best approach to solve these circuits. For example, the above circuit will have only one node equation for node at point E with a voltage vo : vo − v i vo − 0 vo − 0 + − β∆iB + =0 rπ ro RE Because of the controlled source, we need to write an “auxiliary” equation relating the control current (∆iB ) to node voltages: ∆iB = vi − v o rπ

Unless RE is very small (tens of Ω), the fraction in the denominator is quite small compared to 1 and Av ≈ 1. To ﬁnd the input impedance, we calculate ii by KCL: ii = i1 + ∆iB = vi − v o vi + RB rπ 124

ECE65 Lecture Notes (F. Najmabadi), Winter 2006

the output resistance is denoted by Ro (see ﬁgure). the load can be placed in parallel to RE . (and. Winter 2006
125
.
VCC R1 vi Cc vo R2 RE = RL
VCC R1 vi Cc vo R2 RE RL
RE is the Load
Separate Load
vi
Cc
B
∆i
B
rπ
E
β∆ i ro
B
vo RE
vi
Cc
B
∆i
B
rπ
E
β∆ i ro
B
vo
RB
RB
RE
RL
C
Ro
C
R’ o
Alternatively. Najmabadi). Such a circuit is in fact the ﬁrst stage of the 741 OpAmp. The output resistance of the circuit is Ro as is shown in the circuit model. RE . This is usually the case when values of Ro and Ai (current gain) is quoted in electronic text books. we have ii = vi /RB or Ri ≡ vi = RB ii
Note that RB is the combination of our biasing resistors R1 and R2 . itself. In this case. the input resistance of the emitter follower circuit becomes: Ri ≡ vi = RB ii [rπ + (RE ro )(1 + β)]
and it is quite large (hundreds of kΩ to several MΩ) for RB → ∞. In this case. BJT sees a resistance of RE RL . we cannot use vo ≈ vi . is the load. if we want the load not to aﬀect the emitter follower circuit. we should use RL to be much
ECE65 Lecture Notes (F. Obviously. The output resistance of the common collector ampliﬁer (in fact for all transistor ampliﬁers) is somewhat complicated because the load can be conﬁgured in two ways (see ﬁgure): First. Ri large). the input resistance of the emitter follower circuit will become large. This is the case when the common collector is used as a “current ampliﬁer” to raise the power level and to drive the load. therefore RB → ∞).Since vo ≈ vi . This is done when the common collector ampliﬁer is used as a buﬀer (Av ≈ 1. Using the full expression for vo from above. With alternative biasing schemes which do not require R1 and R2 . For this circuit.

As such. Najmabadi). ii ≈ vi /RB and Ai ≡ io RB = ii RE
vi Cc
B
∆i
B
rπ
E
β∆ iB RE
iT vT
RB
ro
We can calculate Ro . (1 + β)(ro ) Ro ≈
(ro ) rπ rπ rπ = ≈ = re (1 + β)(ro ) (1 + β) β
As mentioned above. the output resistance can be found by killing the source (short vi ) and ﬁnding the Thevenin resistance of the two-terminal network (using a test voltage source). We can use our previous results by noting that we can replace ro and RE with ro = ro RE which results in a circuit similar to the case with no RL . the expression for Ro simpliﬁes to
Since. little current ﬂows in RL which is ﬁne because we are using this conﬁguration as a buﬀer and not to amplify the current and power.e. When RE is the load. when RE is the load the common collector is used as a “current ampliﬁer” to raise the current and power levels . the output resistance when an additional load is attached to the circuit (i. In this case. Ro has a similar expression as RO if we replace ro withro :
ECE65 Lecture Notes (F. Therefore.larger than RE .. value of Ro or Ai does not have much use. Winter 2006
+ −
C
R’ o vi Cc
B
∆i
B
rπ
E
β∆ iB
iT vT
RB
r’ o
+ −
C
R’ o
126
. This can be seen by checking the current gain in this ampliﬁer: io = vo /RE . RE is not the load) with a similar procedure: we need to ﬁnd the Thevenin resistance of the two-terminal network (using a test voltage source). vT KCL: iT = −∆iB + − β∆iB ro KVL (outside loop): − rπ ∆iB = vT
vi Cc
B
∆i
B
rπ
E
β∆ iB
iT vT
RB
ro
+ −
C
Ro
Substituting for ∆iB from the 2nd equation in the ﬁrst and rearranging terms we get: Ro ≡ vT (ro ) rπ = iT (1 + β)(ro ) + rπ rπ .

the general properties of the common collector ampliﬁer (emitter follower) include a voltage gain of unity (Av ≈ 1). In most cases. In this case. Winter 2006
127
. CE layers become important and a high-frequency smallsignal model for BJT should be used for analysis. at the ﬁrst stage of an ampliﬁer to provide very large input resistance (such in 741 OpAmp). we have set up a high pass ﬁlter in the input part of the circuit (combination of the coupling capacitor and the input resistance of the ampliﬁer). (1 + β)(ro )
In summary. This combination introduces a lower cut-oﬀ frequency for our ampliﬁer which is the same as the cut-oﬀ frequency of the high-pass ﬁlter: ωl = 2π fl = 1 Ri Cc
Lastly. The coupling capacitor results in a lower cut-oﬀ frequency for the transistor ampliﬁers. The common collector ampliﬁer can be also used as the last stage of some ampliﬁer system to amplify the current (and thus. vi = vi .Ro ≡
vT (ro ) rπ = iT (1 + β)(ro ) + rπ rπ (unless we choose a small value for RE ) and Ro ≈ re
In most circuits. This circuit can be used as buﬀer for matching impedance. our analysis indicates that the ampliﬁer has no upper cut-oﬀ frequency (which is not true). the capacitance between BE . Consider our general model for any ampliﬁer circuit.
Vi Cc Ro + V’ i − Ri Io + AVi Vo − ZL
+ −
+ −
Voltage Amplifier Model
When we account for impedance of the capacitor. You will see these models in upper division
ECE65 Lecture Notes (F. power) and drive a load. In order to ﬁnd the cut-oﬀ frequency. BC. At high frequencies. Najmabadi). Ro is small: Ro = re and current gain can be substantial: Ai = RB /RE . RE is the load. If we assume that coupling capacitor is short circuit (similar to our AC analysis of BJT ampliﬁer). however. we need to repeat the above analysis and include the coupling capacitor impedance in the calculation. a very large input resistance Ri ≈ RB (and can be made much larger with alternate biasing schemes). This is not a correct assumption at low frequencies. we have neglected the impact of the coupling capacitor in the circuit (assumed it was a short circuit). the impact of the coupling capacitor and the lower cut-oﬀ frequency can be deduced be examining the ampliﬁer circuit model. our small signal model is a low-frequency model. As such. Impact of Coupling Capacitor: Up to now.

The solution to include an emitter resistance and use a “bypass” capacitor to short it out for AC signals as is shown. these capacitances results in ampliﬁer gain to drop at high frequencies. the circuit conﬁguration as is shown has as a poor bias. For this circuit. or 2) load is placed in parallel to RC . We need to include RE for good biasing (DC signals) and eliminate it for AC signals. Najmabadi). AC analysis: To start the analysis. common emitter ampliﬁer. Examination of the circuit shows that:
vi Cc
B
∆i
B
C
β∆ iB
vo
vi = rπ ∆iB Av ≡
β vo = − (RC vi rπ Ri = R B r π
vo = −(RC
ro ) β∆iB β RC RC = − rπ re
RB
rπ
ro
RC
ro ) ≈ −
E
Ro R’ o
The negative sign in Av indicates 180◦ phase shift between input and output. the strength of
ECE65 Lecture Notes (F. As with the emitter follower circuit. so your simulation should show the upper cut-oﬀ frequency for BJT ampliﬁers. we see that if vi = 0 (killing the source). The bias point currents and voltages can be found using procedure of pages 110-112. the load can be conﬁgured in two ways: 1) RC is the load. As such. Winter 2006
128
. ∆iB = 0. Analysis of this circuit is straightforward. We see that emitter is now common between input and output AC signals (thus. combine R1 and R2 into RB and replace the BJT with its small signal model. In this case.
VCC VCC
R1 Cc
RC vo vi
R1 Cc
RC vo
vi
R2
R2
RE
Cb
Poor Bias
Good Bias using a by−pass capacitor
For this new circuit and with the capacitors open circuit. The output resistance can be found by killing the source (short vi ) and ﬁnding the Thevenin resistance of the two-terminal network. Basically.courses. we kill all DC sources. this circuit is the same as our good biasing circuit of page 110. Common Emitter Ampliﬁer DC analysis: Recall that an emitter resistor is necessary to provide stability of the bias point. The circuit has a large voltage gain but has a medium value for input resistance. PSpice includes a high-frequency model for BJT.

Therefore. Winter 2006
129
. i. Ro = r o Ro = R C ro
Lower cut-oﬀ frequency: Both the coupling and bypass capacitors contribute to setting the lower cut-oﬀ frequency for this ampliﬁer. RE ≈ re .the dependent current source would be zero and this element would become an open circuit.
where RE ≡ RE Note that usually RE
In the case when these two frequencies are far apart. the cut-oﬀ frequency should be found from simulations. ωl (bypass) ωl (coupling) ωl (coupling) ωl (bypass) → → 1 Ri Cc 1 ωl = 2π fl = RE Cb ωl = 2π fl =
When the two frequencies are close to each other. therefore. both act as a high-pass ﬁlter with: ωl (coupling) = 2π fl = 1 Ri Cc 1 ωl (bypass) = 2π fl = RE Cb re re and. An approximate formula for the cut-oﬀ frequency (accurate within a factor of two and exact at the limits) is: ωl = 2π fl = 1 1 + Ri Cc RE Cb
ECE65 Lecture Notes (F. Najmabadi).e.. there is no exact analytical formulas. the cut-oﬀ frequency of the ampliﬁer is set by the “larger” cut-oﬀ frequency.

The solution is to use a by-pass capacitor as is shown. both the coupling and bypass capacitors contribute to setting the lower cut-oﬀ frequency for this ampliﬁer. Similarly we ﬁnd that an approximate formula for the cut-oﬀ frequency (accurate within a factor of two and exact at the limits) is: ωl = 2π fl = 1 1 + Ri Cc RE Cb (RE1 + re )
where RE ≡ RE2
ECE65 Lecture Notes (F. The AC signal sees an emitter resistance of RE1 while for DC signal the emitter resistance is the larger value of RE = RE1 + RE2 . Obviously formulas for common emitter ampliﬁer with emitter resistance can be applied here by replacing RE with RE1 as in deriving the ampliﬁer gain. VE = RE IE > 1 V cannot be fulﬁlled.A Possible Biasing Problem: The gain of the common emitter ampliﬁer with the emitter resistance is approximately RC /RE . we “short” the bypass capacitor so RE2 is eﬀectively removed from the circuit. however. Najmabadi). For cases when a high gain (gains larger than 5-10) is needed. Winter 2006
133
. and input and output impedances.
VCC
R1 Cc
RC vo
vi
R2
R E1
R E2
Cb
The addition of by-pass capacitor. modiﬁes the lower cut-oﬀ frequency of the circuit. Similar to a regular common emitter ampliﬁer with no emitter resistance. RE may be become so small that the necessary good biasing condition.

. e.The overall gain of the two-stage ampliﬁer is then Av = Av1 × Av2 = 4. The ﬁnd the lower cut-oﬀ frequency of the two-stage ampliﬁer is 6.
ECE65 Lecture Notes (F. The drawback of the Example 6 circuit is that the bias circuit is more complicated and harder to design. the overall cut-oﬀ frequency of the circuit is lower. use a large RE2 . use a large capacitor for coupling the two stages.9 kΩ. Some of these issues can be resolved by design. the second-stage does not load the ﬁrst stage and the overall gain is higher. The two-stage ampliﬁer of Example 6 has many advantages over that of Example 5. Najmabadi).5 Hz. The input resistance of the two-stage ampliﬁer is the input resistance of the ﬁrst-stage (Tr1). It has three less elements. Also because of the absence of a coupling capacitor between the two-stages.g. Because of the absence of bias resistors.. Ri = 9. Winter 2006
145
. etc.